Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long sy, a, b;
int n;
long long L[200005], R[200005];
long long smt[800005];
long long x, y;
void build(int idx, int l, int r) {
  if (l == r) {
    smt[idx] = R[l] - L[l];
    return;
  }
  int m = (l + r) / 2;
  build(idx << 1, l, m);
  build(idx << 1 | 1, m + 1, r);
  smt[idx] = smt[idx << 1] + smt[idx << 1 | 1];
}
long long query(int idx, int l, int r, int s, int t) {
  if (r < s || l > t) return 0;
  if (l >= s && r <= t) return smt[idx];
  int m = (l + r) / 2;
  return query(idx << 1, l, m, s, t) + query(idx << 1 | 1, m + 1, r, s, t);
}
int main() {
  scanf("%lld%lld%lld", &sy, &a, &b);
  sy = -sy;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%lld%lld", &L[i], &R[i]);
  }
  build(1, 1, n);
  int q;
  scanf("%d", &q);
  while (q--) {
    scanf("%lld%lld", &x, &y);
    int LL, RR;
    int lb = 0, ub = n + 1;
    while (ub > lb + 1) {
      int m = (ub + lb) / 2;
      if ((y + sy) * R[m] - sy * x <= y * a)
        lb = m;
      else
        ub = m;
    }
    LL = ub;
    lb = 0, ub = n + 1;
    while (ub > lb + 1) {
      int m = (ub + lb) / 2;
      if ((y + sy) * L[m] - sy * x >= y * b)
        ub = m;
      else
        lb = m;
    }
    RR = lb;
    if (RR < LL) {
      printf("0\n");
      continue;
    }
    if (RR == LL) {
      long long right = min(b * y, (y + sy) * R[RR] - sy * x);
      long long left = max(a * y, (y + sy) * L[LL] - sy * x);
      printf("%.8f\n", (right - left) * 1.0 / y);
      continue;
    }
    if (RR == LL + 1) {
      long long right1 = min(b * y, (y + sy) * R[RR] - sy * x);
      long long left1 = (y + sy) * L[RR] - sy * x;
      long long right2 = (y + sy) * R[LL] - sy * x;
      long long left2 = max(a * y, (y + sy) * L[LL] - sy * x);
      printf("%.8f\n", (right1 - left1 + right2 - left2) * 1.0 / y);
      continue;
    }
    long long tot = query(1, 1, n, LL + 1, RR - 1);
    long long right1 = min(b * y, (y + sy) * R[RR] - sy * x);
    long long left1 = (y + sy) * L[RR] - sy * x;
    long long right2 = (y + sy) * R[LL] - sy * x;
    long long left2 = max(a * y, (y + sy) * L[LL] - sy * x);
    printf("%.8f\n",
           (right1 - left1 + right2 - left2 + (sy + y) * tot) * 1.0 / y);
  }
  return 0;
}